Find the gradient of $f(x, y, z) = z^2y^2 - e^x$. $\nabla f = ($ $,$ $,$ $)$
Solution: The gradient of a scalar field is all its partial derivatives put together into a vector. For a 3D scalar field, this looks like $\nabla f = (f_x, f_y, f_z)$. Let's find $f_x$, $f_y$, and $f_z$. $\begin{aligned} f_x &= \dfrac{\partial}{\partial x} \left[ z^2y^2 - e^x \right] \\ \\ &= -e^x \\ \\ f_y &= \dfrac{\partial}{\partial y} \left[ z^2y^2 - e^x \right] \\ \\ &= 2yz^2 \\ \\ f_z &= \dfrac{\partial}{\partial z} \left[ z^2y^2 - e^x \right] \\ \\ &= 2zy^2 \end{aligned}$ The gradient of $f$ is: $\nabla f = \left( -e^x, 2yz^2, 2zy^2 \right)$